HPSC PGT Mathematics Mock Test - 7 - HPSC TGT/PGT MCQ

The correct answer is Bronze.

Key Points

• Vinesh Phogat has won the bronze medal in the ongoing World Wrestling Championships in Belgrade.
• She has become the first Indian woman wrestler to win two world championship medals.
• She won the bronze medal in the 53 kg freestyle event by defeating Sweden's Jonna Malmgren 8–0.

Additional Information

• ​Vinesh won three gold in Commonwealth Games.
• She won a gold medal in wrestling in the 48 kg category at the 2014 Commonwealth Games, and in the 50 kg category at the 2018 Commonwealth Games.
• After this, Vinesh won gold in the 53 kg weight category in the Commonwealth Games this year as well.

International Yoga Day is celebrated on 21^st June every year.

Similarly;

Kargil Victory day is commemorated on 26^th July every year.

Hence, "Kargil Victory Day : 26^th July" is the correct answer.

Important Points

GIVEN:

The average of 40 numbers = 25

CONCEPT:

Basic Average concept.

FORMULA USED:

Average = Sum of all numbers/Total numbers

CALCULATION:

Sum of numbers = 40 × 25 = 1000

23, 25 and 41 are mistakenly written as 32, 52 and 15 respectively.

⇒ Change in sum = (23 + 25 + 41) – (32 + 52 + 15)

= 89 – 99

= -10

Actual sum of numbers will be 10 less.

⇒ Sum of numbers = 1000 – 10 = 990

∴ Correct average = 990/40 = 24.75

The pattern followed is, 

Logic: For all the other option sum of place value is 14 except “DJA”

Hence, “DJA” is the correct answer.

Change in behaviour is the common sign of a child in stress. A child in stress shows aggressive behaviour and acts irritably.

It is distributive law.

Given, function f : R→R such that f(x) = x², 
Let A and B be two sets of real numbers.
Let x₁, x₂ ∈ A such that f(x₁) = f(x₂).
⇒1 + (x₁)² = 1 + (x₂)²
​⇒ (x₁)² − (x₂)² = 0
⇒(x₁ − x₂)(x₁ + x₂)=0
⇒ x₁ = ± x₂. 
Thus f(x₁) = f(x₂) does not imply that x₁ = x₂.
For instance, f(1) = f(−1) = 1, i.e. , two elements (1, -1) of A have the same image in B. So, f is many-one function.

We have, zk = ωk where ω  



(by De Movire's theorem)
= cos π + i sin π = -1

On rationalizing,
 Lt x-->0 (1-Cosx)/[√(1+x) +1]
= lt x-->0 2sin²(x/2)[√(1+x)+1] /x
= lt x-->0 [2sin(x/2)/x].sinx/2 [√(1+x) +1 ]
= 2.1/2 .0[1+1]=0

S_p = Sq
⇒  p/2(2a+(p−1)d) = q/2(2a+(q−1)d)
⇒ p(2a+(p−1)d) = q(2a+(q−1)d)
⇒ 2ap + p²d − pd = 2aq + q²d − qd
⇒ 2a(p−q) + (p+q)(p−q)d − d(p−q) = 0
⇒ (p−q)[2a + (p+q)d − d] = 0
⇒ 2a + (p+q)d − d = 0
⇒ 2a + ((p+q) − 1)d = 0
⇒ S_p+q = 0

z lies inside or on the ellipse.

Clearly the minimum distance of z from the given point 4 is 1 and maximum distance is 9

(i) xRx ⇔ 2x² - 3x.x + x²  

∴ R is reflexive
(ii) For x = 1, y= 2; 2x² - 3xy + y² = 0
∴ 1 R 2 but 2.2² - 3.2.1 + 1² = 3 ≠ 0.
So, 2 is not R-related to 1.
∴ R is not symmetric. 

Equation of tangent at (x1 , y1) to the circle x² + y² =25 is given by,
xx1 + yy1 = 25. Now comparing it with 3x+4y = 25
We get required point of contact (3,4)

P(A∪B) = P(A) = P(A) + P(B) - P(A∩B)
P(B) = 2/3
 = P(B) - P(A∩B) = 

We have :

If A = Ф that means A does not contain any element i.e., n = 0.
Now, number of elements in a power set is 2ⁿ.
∴, n[P(A)] = 2⁰ = 1
Therefore P(A) contains 1 element.

T_(r+1) = ¹¹C_r x^(11-2r) (m/r)^r (-1)^r
= ¹¹C_r x^(11-2r) m^r (-1)^r
Coefficient of x³ = 11 - 2r = 3
8 = 2r 
r = 4
T5 = ¹¹C₄ x³ m⁴ (-1)
Coefficient of x³ = ¹¹C₄ m⁴
= 330 m⁴

We have f: N →I
If x and y are two even natural numbers,

Again if x and y are two odd natural numbers then

Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.

∴ f is onto.
Hence f is one one and onto both.

For every square matrix (A – A’) is always skew – symmetric.

Obviously, every differentiable function is continuous but every continuous function isn't differentiable.

y = log(secx + tanx)
dy/dx = 1/(secx + tanx){(secxtanx) + sec^2x}
= secx(secx + tanx)/(secx + tanx)
= secx

Total no of balls = 9
red balls = 4
yellow balls = 3
green balls = 2
Total no. of arrangements = 9!/(4!*3!*2!)
= 1260

The function is clearly differentiable except possible at x = 2, 3